If n and y are positive integers and 450y = n^3 which of the following

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

31 Oct 2013, 05:49

Kudos

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I finally get it. If you want the cuberoot of 450y to be an integer, you need the primefactors to be cubed.

So the primefactors we have for 450 are 2, 3², 5². To be perfect cubes, we need to multiply by 2², 3, 5 which in this case equals y.

so we get that (2*3²*5²) * (2²*3*5) = 2³*3³*5³. If we'd cuberoot x now, we'd get an integer.

This leads to the conclusion that y / 2²*3*5 = 1 = integer. So answer B.

I tried explaining in simpler words so I'd get it myself, I hope it helps the you other non-math-overbrains.

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

19 Jun 2012, 22:45

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enigma123 wrote:

If n and y are positive integers and 450y = n^3, which of the following must be an integer?

1. $\frac{Y}{3 * 2^2 * 5}$

2. $\frac{Y}{3^2 * 2 * 5}$

3. $\frac{Y}{3 * 2 * 5^2}$

A) None

B) I only

C) II Only

D) III only

E) I, II and III onmly.

The critical point in this question is that n and y are integers.
Since n is an integer, n^3 must be the cube of an integer. So in n^3, all the prime factors of n must be cubed (or have higher powers which are multiples of 3)

Now consider 450y = n^3
450 is not a perfect cube. So whatever is missing in 450, must be provided by y to make a perfect cube.

$450 = 45*10 = 2 * 3^2 * 5^2$

To make a perfect cube, we need at least two 2s, a 3 and a 5. These missing factors must be provided by y. Therefore, $2^2*3*5$ must be a factor of y.
This means $y/2^2*3*5$ must be an integer.
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Bunuel

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

13 Apr 2010, 02:58

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shrivastavarohit wrote:

Moderator,

Can you elaborate more on the solution, seems unfathomable for simple minds like mine

.

How did we deduce that "Minimum value of Y"?

$x$ and $y$ are integers and $450y=x^3$ --> $450y$ equals to cube of an integer. $450y=2*3^2*5^2*y=x^3$. The smallest value of $y$ for which $2*3^2*5^2*y$ is a cube of an integer is when $y=2^2*3*5$. In this case $450y=(2*3^2*5^2)*(2^2*3*5)=(2*3*5)^3$. Of course $y$ can take another values as well, for example $y=2^5*3^4*5^7$ and in this case $450y=(2*3^2*5^2)*(2^5*3^4*5^7)=(2^3*3^2*5^3)^3$, but the smallest value of $y$ is when $y=2^2*3*5$.

You can check the similar problems at:
og-quantitative-91750.html#p704028
division-factor-88388.html#p666722
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shrivastavarohit

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

12 Apr 2010, 23:15

Kudos

Moderator,

Can you elaborate more on the solution, seems unfathomable for simple minds like mine

.

How did we deduce that "Minimum value of Y"?

shrouded1

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

12 Oct 2010, 00:14

Kudos

atomjuggler wrote:

Raising the white flag on this one. I really hope I'm not missing something obvious

Source: GMATPrep Test #1

If n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. $\frac{y}{3*2^2*5}$

II. $\frac{y}{3^2*2*5}$

III. $\frac{y}{3*2*5^2}$

A. None
B. I only
C. II only
D. III only
E. I, II, and III

$450*y=n^3$
Since 450y is a whole cube, in the prime factorization of 450y, all the exponents on the prime factors must be multiples of 3.
$450*y = y * 3^2 * 5^2 * 2$
So y must have atleast $3, 5, 2^2$ as prime factors. y must be of the form form 3x5x4xA where A is also another perfect cube.
By this logic the answer is (I) only or B, as in all cases y/(3x5x4) has to be an integer.
(ii) or (iii) cannot be true if A=1

Answer is (b)
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

22 May 2014, 11:59

I could understand the options

But tell y my alternate approach is wrong
450y = x^3

X and Y are integer
450 = 2*3^2*5^2

X=((2*3^2*5^2)y)/x^2
Note: as x is an integer
if x = 2*3^2*5^2

X = y/ (2*3^2*5^2)

And its option c).

Please correct me.

Thanks in advance.

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

22 May 2014, 13:11

Kudos

Hi rrsnathan,

Your approach is wrong because you are assuming 1 fixed value of $x = 2*3^2*5^2$

But question explicitly asks which option MUST be an integer always. In your assumption, Option C comes out to be an integer but that may not be true in other cases.

So, Right method to solve it is:

$450y = x^3$

$(2*3^2*5^2) * y = x^3$

For LHS to be a cube (As R.H.S is a cube),

$y = 2^2*3*5*k^3$ (where k is a positive integer)

Hence,
Option (b) is always correct

Rgds,
Rajat

KarishmaB

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

22 May 2014, 20:25

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rrsnathan wrote:

This is incorrect. Note that in the denominator, you have x^2, not x. You are assuming $x = 2*3^2*5^2$ and putting it as it is in the denominator. Also note that since you have the same x on the left hand side, if you put $x = 2*3^2*5^2$, you get

$2*3^2*5^2 = (2*3^2*5^2)*y/(2*3^2*5^2)^2$
i.e. $y = (2*3^2*5^2)^2$

Basically, you have assumed a value of x and got a value of y. There is no reason for you to assume the value of x as $2*3^2*5^2$.

The point here is that y must have some values such that when it multiplies 450, it gives a perfect cube. This is the method discussed above in my post.
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Bunuel

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

23 May 2014, 10:53

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jakexix

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

30 Jun 2017, 00:04

Cant we take higher values of N and Y so that it satisfies the equation,the all 3 possiblities are correct and answer is E
eg:
y= 5^4*3^4*2^5
N=5^2*3^2*2^2

Bunuel

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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]

30 Jun 2017, 00:18

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jakexix wrote:

Cant we take higher values of N and Y so that it satisfies the equation,the all 3 possiblities are correct and answer is E
eg:
y= 5^4*3^4*2^5
N=5^2*3^2*2^2

The question asks which of the following MUST be an integer, not COULD be an integer. If y = 2^2*3*5, which is the least possible value of y, then only B is true. Please re-read the thread and follow the links to similar questions to understand the concept better.
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